Question

A string, fixed at both ends, vibrates in a resonant mode with a separation of 2⋅0 cm between the consecutive nodes. For the next higher resonant frequency, this separation is reduced to 1⋅6 cm. Find the length of the string.

Answer 1

Given:
Separation between two consecutive nodes when the string vibrates in resonant mode = 2.0 cm
Let there be 'n' loops and
\[\lambda\] be the wavelength.
∴
\[\lambda\] = \[2 \times Separation  between  the  consecutive  nodes\]

\[\lambda_1  = 2 \times 2 = 4  \text{ cm }\]

\[\lambda_2  = 2 \times 1 . 6 = 3 . 2  cm\]
Length of the wire is L.

In the first case:

\[L = \left( \frac{n \lambda_1}{2} \right)\] 

In the second case:

\[L = \left( n + 1 \right)\frac{\lambda_2}{2}\] 

\[ \Rightarrow \frac{n \lambda_1}{2} = \left( n + 1 \right)  \frac{\lambda_2}{2}\] 

\[ \Rightarrow n \times 4 = \left( n + 1 \right)\left( 3 . 2 \right)\] 

\[ \Rightarrow 4n - 3 . 2n = 3 . 2\] 

\[ \Rightarrow 0 . 8  n = 3 . 2\] 

\[ \Rightarrow n = 4
\text{ ∴ length of the string,}\]
\[L = \frac{\left( n \lambda_1 \right)}{2} = \frac{\left( 4 \times 4 \right)}{2} = 8  \text{ cm }\]